Cho đa thức f(x) = $x^{6}$ – $2021x^{5}$ – $2021x^{4}$ – $2021x^{3}$ – $2021x^{2}$ – $2021x$ + 2021
Tính f(2020)
Cho đa thức f(x) = $x^{6}$ – $2021x^{5}$ – $2021x^{4}$ – $2021x^{3}$ – $2021x^{2}$ – $2021x$ + 2021
Tính f(2020)
Tham khảo
Sửa:`f(x)=x^6-2021x^5+2021x^4-2021x^3+2021x^2-2021x+2021`
`⇒f(2020)=2020^6-2021.2020^5+2021.2020^4-2021.2020^3+2021.2020^2-2021.2020+2021`
`⇒f(2020)=2020^6-2020^6-2020^5+2020^5+2020^4-2020^4-2020^3+2020^3+2020^2-2020^2-2020+2020+1`
`⇒f(2020)=(2020^6-2020^6)+(-2020^5+2020^5)+(2020^4-2020^4)+(-2020^3+2020^3)+(2020^2-2020^2)+(-2020+2020)+1`
`⇒f(2020)=1`
`\text{©CBT}`
Đáp án: $f(2020)=1$
Giải thích các bước giải:
Sửa lại đề:
$f(x)=x^6-2021x^5+2021x^4-2021x^3+2021x^2-2021x+2021$
$f(2020)=2020^6-2021.2020^5+2021.2020^4-2021.2020^3+2021.2020^2-2021.2020+2021$
$=2020^6-2020.2020^5-2020^5+2020.2020^4+2020^4-2020.2020^3-2020^3+2020.2020^2+2020^2-2020.2020-2020+2020+1$
$=2020^6-2020^6-2020^5+2020^5+2020^4-2020^4-2020^3+2020^3+2020^2-2020^2-2020+2020+1$
$=1$